Problem: Simplify the following expression and state the condition under which the simplification is valid: $q = \dfrac{x^2 + 11x + 28}{x^2 + 13x + 42}$
First factor the expressions in the numerator and denominator. $ \dfrac{x^2 + 11x + 28}{x^2 + 13x + 42} = \dfrac{(x + 4)(x + 7)}{(x + 6)(x + 7)} $ Notice that the term $(x + 7)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(x + 7)$ gives: $q = \dfrac{x + 4}{x + 6}$ Since we divided by $(x + 7)$, $x \neq -7$. $q = \dfrac{x + 4}{x + 6}; \space x \neq -7$